Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 329 / 2

Let $z$ increase downward. The slender viscous thread is locally in uniaxial extension. The transverse stress equals the ambient pressure, so the Trouton ratio gives the excess axial stress ${\sigma }_{zz}+p_e=3\mu w_z$. Mass conservation and axial force balance therefore give

In steady flow $Aw=Q$. Dividing the momentum equation by $A=Q/w$ givesWith $w=\hat{w}W$ and $z=\hat{z}Z$, chooseThe dimensionless equation becomesTreating $(W^{\prime }{)}^2$ as a function of $W$ and using an integrating factor yieldsBecause $A=Q/(\hat{w}W)$, the area sketches are the reciprocals of the functions $W$ found below. The dimensional vertical deviatoric stress is

For $C=0$, the increasing branch with $W(0)=0$ satisfies $W^{\prime }=\sqrt{2W}$, henceThe thread thins algebraically, and the tensile vertical stress is proportional to $W^{\prime }=Z$.

For $C=1/2$,soThe thread thins exponentially for large $Z$, and its tensile stress is proportional to $\sinh Z$.

For $C=-1/2$,so the branch emerging from zero isThe area decreases until $Z=\pi$, where $W=2$ and the vertical deviatoric stress, proportional to $W^{\prime }=\sin Z$, vanishes. A formal continuation beyond that point has compressive stress and thickens until the model reaches another zero of $W$; unlike the first two cases, it does not describe indefinite monotone drawing.

The broad faces have curvature zero to leading order, so their stress boundary condition gives ${\sigma }_{xx}=-p_e$. The semicircular edges have curvature $1/h$, giving ${\sigma }_{yy}=-p_e-\gamma /h$. Incompressible flow gives $u_x+v_y+w_z=0$; uniform transverse normal stresses imply uniform transverse extension rates, and eliminating them from the Newtonian fluid stress tensor yields

At an edge, the kinematic boundary condition balances axial advection of $b$, lateral strain, and capillary retraction. This givesThe equation $hbw=Q$ expresses conservation of volume flux, whilestates that the total axial tension is constant. DefineThen $[\Gamma ]=L^{-2}$ and $[T]=L^{-1}$, and substitution of $h=Q/(bw)$ givesSubtracting these logarithmic-derivative equations givesThe remaining width equation isThus, for $T\ne 0$,while the continuous $T=0$ limit is $1/b=1/b_0+\Gamma z/6$.

If $\Gamma =0$, then $b/b_0=e^{-Tz/6}$, $w/w(0)=e^{Tz/3}$, and mass conservation gives $h/h(0)=e^{-Tz/6}$. Therefore a prescribed thinning ratio $R=h(L)/h(0)$ requires